Question

Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (4He) from gold-197 (197 Au). See the figure below. The energy of the incoming helium nucleus was 9.90 x 10-13 J, and the masses of the helium and gold nuclei were 6.68 x 10-27 and 3.29 x 10-25 kg, respectively (note that their mass ratio is 4 to 197). If a helium nucleus scatters to an angle of 120° during an elastic collision with a gold nucleus, calculate the helium nucleus's final speed. m/s Calculate the final velocity (magnitude and direction) of the gold nucleus. (Assume the positive x direction is the direction in which the helium nucleus is initially traveling, and that it scatters 120° clockwise from the +x-axis.) magnitude ( m/s direction (counterclockwise from the +x-axis) 120 Before collision After collision

Answer 1

& conservation of Energy is, Mhe vi he = m he vite + Mau Ve Au ... 1. 2 2 1. = 1. » conservation of momentum in x-direction, mhe Vine = Mhe Vf he coso + Mau Vs Au coso... conservation of momentum in y-direction, mie Us he sino = mau Vf Au sind .... ③ At this point, square of een ♡ is, mu v Au sinho = m. the us a me sinto ... 4 by Rearranging Eqn ②,

mine he cost o mu v Au cos d = mil the vil he .am the Nike Vs He costot O by adding Eqn 6 & 6, mau viou (cos? ¢ + sin o) = mãe vs He (sin’o + cos20) + me he vue - am the Vine the cosce cos'è + sin?d=1 4 cos'o tsin?o=1 ---- me vs Are = mine vine + mume vitine - and the Vine is he coso ----

from an oa 2 2 2 m Au vt Au P " He VfHe + Vine - 2 Vine Vene oso) He MHE MAU (Wil the vihe) Mhe man (with by Rearranging , viure (vt mate) + vine ('- max. ) - eve ve Uste cose не mhe - Insenting the ratio mau for value ' & the mHe 14 $. V. He = 2 kihe - = 2.96 X 10 the angle o, we have the quadratic, vthe li+197 ) + vitte (1-194)- 2 Vi he Vere COS 120* og CS0.3) viie + Vine Vithe -(45.3) v1He =0 8. Kite = mne ve me vine dikime = 24.909 210 oo VPhe = 1.72 x 107 mis Inserting vil he & vine into Eqn ☺ , (50.3) vite +(1.72 x10%) vs He -(48.3) ( 2.96 x 104)=0 :: Vue = 1066 x107 mis using Basic definition, of kinetic Energy, by Rewrite Ean , 2 x 9.9*10-13 92 6.68 x 10 мне 1. - 1.

. . Q Kite = mue of the than Vs Au 8.219.9 x 10-13) -(6.68x10-29) (1.66 x 10°)" + (3.29 x10-25) vi alle | 8 V Au = 8,914 2 lo- 3.29 X 107 10-25 1 X for sin & • Venge = 1.62 * 106 mice finally Ean ③ may be solved & values inserted to obtain, sind = mie Vf he sino 1. MAU Vt au maz Vf Au 1.62 X 106 sind = (mue) ( Vf He ) sino :. son = 662) (1366X10%) sin 120* i sin a = 0.180 ;. $ = 10.369° Vshe = 1.66 x 107 mis VEAU = 1,62 x 106 mis (magnarude) direction = 10.369° (counter clockwise from = x-axis)